Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{t^2 - 4t}{t^2 + 4t - 21} \div \dfrac{-8t + 32}{t + 7} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{t^2 - 4t}{t^2 + 4t - 21} \times \dfrac{t + 7}{-8t + 32} $ First factor the quadratic. $y = \dfrac{t^2 - 4t}{(t + 7)(t - 3)} \times \dfrac{t + 7}{-8t + 32} $ Then factor out any other terms. $y = \dfrac{t(t - 4)}{(t + 7)(t - 3)} \times \dfrac{t + 7}{-8(t - 4)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ t(t - 4) \times (t + 7) } { (t + 7)(t - 3) \times -8(t - 4) } $ $y = \dfrac{ t(t - 4)(t + 7)}{ -8(t + 7)(t - 3)(t - 4)} $ Notice that $(t - 4)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ t(t - 4)\cancel{(t + 7)}}{ -8\cancel{(t + 7)}(t - 3)(t - 4)} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $y = \dfrac{ t\cancel{(t - 4)}\cancel{(t + 7)}}{ -8\cancel{(t + 7)}(t - 3)\cancel{(t - 4)}} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $y = \dfrac{t}{-8(t - 3)} $ $y = \dfrac{-t}{8(t - 3)} ; \space t \neq -7 ; \space t \neq 4 $